3-4 solving systems of linear equations in 3 variables
DESCRIPTION
3-4 Solving Systems of Linear Equations in 3 Variables. Algebra 2 Textbook. Terms to remember. Visualizing what this means. Using elimination to solve. STEP 1. Rewrite the system as a linear system in two variables. 4 x + 2 y + 3 z = 1. Add 2 times Equation 3. - PowerPoint PPT PresentationTRANSCRIPT
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3-4 Solving Systems of Linear Equations in 3 Variables
Algebra 2 Textbook
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Terms to remember
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Visualizing what this means
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Using elimination to solve
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EXAMPLE 1
Use the elimination method
Solve the system. 4x + 2y + 3z = 1 Equation 1
2x – 3y + 5z = – 14 Equation 2
6x – y + 4z = – 1 Equation 3
SOLUTION
STEP 1Rewrite the system as a linear system in two variables.
4x + 2y + 3z = 1
12x – 2y + 8z = – 2
Add 2 times Equation 3
to Equation 1.
16x + 11z = – 1 New Equation A
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EXAMPLE 1
2x – 3y + 5z = – 14
– 18x + 3y – 12z = 3
Add – 3 times Equation 3to Equation 2.
– 16x – 7z = – 11 New Equation B
STEP 2 Solve the new linear system for both of its variables.
16x + 11z = –1 Add new Equation A
and new Equation B.– 16x – 7z = –11
4z = –12
z = – 3 Solve for z.
x = 2 Substitute into new Equation A or B to find x.
Use the elimination method
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6x – y + 4z = – 1
EXAMPLE 1
Use the elimination method
STEP 3Substitute x = 2 and z = – 3 into an original equation and solve for y.
Write original Equation 3.
6(2) – y + 4(– 3) = – 1 Substitute 2 for x and –3 for z.
y = 1 Solve for y.
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GUIDED PRACTICE for Examples 1, 2 and 3
Solve the system.
1. 3x + y – 2z = 10
6x – 2y + z = – 2
x + 4y + 3z = 7
ANSWER (1, 3, – 2)
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EXAMPLE 2
Solve a three-variable system with no solution
Solve the system. x + y + z = 3 Equation 1
4x + 4y + 4z = 7 Equation 2
3x – y + 2z = 5 Equation 3SOLUTION
When you multiply Equation 1 by – 4 and add the result to Equation 2, you obtain a false equation. Add – 4 times Equation 1
to Equation 2.– 4x – 4y – 4z = – 12
4x + 4y + 4z = 7
0 = – 5
New Equation A
Because you obtain a false equation, you can conclude that the original system has no solution.
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EXAMPLE 3
Solve a three-variable system with many solutions
Solve the system. x + y + z = 4 Equation 1
x + y – z = 4 Equation 2
3x + 3y + z = 12 Equation 3
SOLUTION
STEP 1 Rewrite the system as a linear system in two variables.
Add Equation 1
to Equation 2.
x + y + z = 4
x + y – z = 4
2x + 2y = 8 New Equation A
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EXAMPLE 3
Solve a three-variable system with many solutions
x + y – z = 4 Add Equation 2
3x + 3y + z = 12 to Equation 3.
4x + 4y = 16 New Equation B
Solve the new linear system for both of its variables.
STEP 2
Add –2 times new Equation A
to new Equation B.
Because you obtain the identity 0 = 0, the system has infinitely many solutions.
– 4x – 4y = – 16
4x + 4y = 16
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EXAMPLE 3
Solve a three-variable system with many solutions
STEP 3 Describe the solutions of the system. One way to do this is to divide new Equation 1 by 2 to get x + y = 4, or y = – x + 4. Substituting this into original Equation 1 produces z = 0. So, any ordered triple of the form (x, – x + 4, 0) is a solution of the system.
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GUIDED PRACTICE for Examples 1, 2 and 3
2. x + y – z = 22x + 2y – 2z = 65x + y – 3z = 8
ANSWER no solution
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GUIDED PRACTICE for Examples 1, 2 and 3
3. x + y + z = 3x + y – z = 3
2x + 2y + z = 6
ANSWERInfinitely many solutions
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3.20
4.80
3.68
Step 1 Write the Equations
Step 2 Use substitution to solve the system
Step 3 Write the answers down with correct labels
Eq. 1 p + c = 100Eq. 2 3.2p + 4.8c = 368
Solve Eq. 1 for p p = 100-c Substitute 100-c in for p in Eq. 2 3.2(100-c) + 4.8c = 368Use distribute property to simplify 320 – 3.20c + 4.8c = 368Combine like terms 320 + 1.6c = 368Subtract 320 from each side 1.6c = 48Solve for c then p by substitution c = 30 p = 100 – 30 = 70
30 pounds of cashews & 70 pounds of peanuts